Physics-
General
Easy

Question

A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at a distance 3R from the centre of the sphere. A spherical cavity of radius R/2 is now made in the sphere as shown in the figure. The sphere with cavity now applies a gravitational force F2 on the same particle. The ratio F2 /F1 is

  1. fraction numerator 9 over denominator 50 end fraction    
  2. fraction numerator 41 over denominator 50 end fraction    
  3. fraction numerator 3 over denominator 25 end fraction    
  4. fraction numerator 22 over denominator 25 end fraction    

hintHint:

Let the particle of mass m be place on A then
F1 = GMm/(2R)^2

The correct answer is: fraction numerator 41 over denominator 50 end fraction

F1=GMm/(2R)^2

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